Problem 23-2 Source:
Problem 7.35 - page 301 Book: Análise de Circuitos em Engenharia - J. David Irwin - 4ª edição - Ed. Pearson - 2013.
In the circuit show in the Figure 23-02.1, the S switch has been closed for a long time. For the time
t = 0 the switch has been opened. Calculate the current i(t) to t ≥ 0.
Figure 23-02.1
Solution of the Problem 23-2
As the S switch remained on for a long time, so in this case the inductor behaves like a short circuit. This way we can calculate the current iL circulating through the inductor in time t = O-.
iL (0-) = 24 / 6 = 4 A
Analyzing the situation when opening the S switch, it is known that the inductor cannot abruptly change the value of the current flowing through it. Then, it is concluded that the value of the current in the inductor, iL(0+) = 4 A.
On the other hand, it is clear from the circuit that it is possible to calculate the equivalent resistance to the right of the inductor. We have the series of resistors 4 + 8 = 12 Ω in parallel with the resistor of 6 Ω. This results in a resistor of 4 Ω. Now just add with the resistor of 4 Ω and find the value of 8 Ω. Now, you should parallel this to the
12 Ω resistor. Thus, the value of the equivalent circuit resistance is calculated when the
S switch is open, which is equal to 24/5 ohms. This way we can calculate the time constant of the circuit.
τ = L / Req = 1 / (24/5) = 5/24 s
Note that for t = 0+, the circuit boils down to the inductor and a resistor. Therefore, we easily conclude that the current value iL when
t → ∞ equals zero as there are no sources of any kind that can keep a current flowing through the circuit. Knowing the value of the time constant, the start current and the end current, we can calculate the circuit response using eq. 23-03 studied in item 4. Substituting for numeric values gives:
iL(t) = 4 e-4,8 t A, t ≥ 0
In the above equation we use the fact that -t / (5/24) = -4.8t.
To find the current i(t), a current divider may be used. However, we will present another technique that can be used: as we know the equation of the current flowing through the inductor, this allows one to calculate the voltage on the inductor; just derive the current and multiply by the inductor inductance value. Doing this we find:
VL(t) = -19.2 e-4.8t V
We must pay attention to the fact that the three resistors to the right of the circuit can be replaced by a resistor of 4 ohms. And adding this value with the resistor of 4 ohms (where it circulates i(t)), we have a
total of 8 ohms as equivalent resistance of this part of the circuit.
Now the voltage VL(t) It is also about this equivalent resistance. So to find the current i(t) just divide these two quantities, or:
i (t) = VL(t) / Req = -19.2 e-4.8t / 8
Performing the calculation, we obtain:
i (t) = -2.4 e-4.8t V, t ≥ 0
When calculating using a current divider, the same value must be found. Note the negative sign in
i(t), as it is in the opposite direction to the current in the inductor. When using the derivative, the negative sign automatically appears by definition. This presented technique is another way to solve this kind of problem.
